Mathematics Asked on January 5, 2022

I don’t know what to do with the following question:

Show that if a Banach space is generated by a countable set with finite dimension then it has finite dimension

Supposedly it’s related to the Baire theorem, but that didn’t help me one bit. In fact it looks totally unrelated:

Theorem (Baire)If $(G_n)_{n geq 1}$ is a sequence of open, dense subsets of the complete metric space $(X,d)$, then the intersection $bigcap_{n geq 1} G_n$ is dense in $X$.

So how would one go solving this?

**Edit** At first I hadn't read the title of your question so I thought that you wanted to prove Baire's theorem and I added an answer based on that. My bad.

However, I do know how to show what you have on your title using Baire's theorem so here's a sketch:

I'll leave to you to check all the details.

The following statement is a direct consequence of Baire's theorem

Theorem.If $(X,d)$ is a complete metric space and $X=bigcup_{n geq 1} X_n$, then $mathrm{int}(overline{X_n}) neq varnothing$ for some $n$.

Now I will use the theorem to show that the algebraic dimension of a Banach space is either finite or uncountable:

Assume that $X$ is a Banach space generated by ${ xi_n : n in mathbb{Z}_{>0}}$. For each $n in mathbb{Z}_{>0}$ let $X_n$ be the space generated by ${xi_1, ldots, xi_n}$. Now it's standard to show that

- $X_n$ is closed
- $mathrm{int}(X_n)= varnothing$

Since $X$ is generated by these $X_n$'s, we have $X=bigcup_{n geq 1} X_n$. Thus, by the theorem above $mathrm{int}(X_n) neq varnothing$ for some $n$, a contradiction.

Answered by Alonso Delfín on January 5, 2022

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